In one process ,4.0g of sulphur was burnt in 48dm³ of oxygen measured at r.t.p, calculate the volume of sulphur dioxide formed at r.t.p
Technically sulfur is S8 but it's simpler to use S and the same amount of SO2 will be formed so let me take the shortcut. This is a limiting reagent (LR) problem and you know that when an amount for more than one of the reactants is given. First we must determine the LR.
S + O2 ==> SO2
MOLS S = 4.0/32.0 = 1/8 = 0.125
mols O = 48 dm^3/24 = 2
If all of the S is used how much O2 will be needed? That's
0.125 x (1 mol O2/1 mol S) = 0.125 x 1/1 = 0.125. Do you have that much O2. Yes, therefore, S is the LR and O2 is the excess reagent (ER). So 0.125 mols S will form 0.125 mols SO2 gas. What volume will that occupy? That will be 0.125 mols SO2 x (24 dm^3/mol) = 0.125 mol x 24 dm^3/mol = ? dm^3